chapter 2 section 2.4 Overfencing-优快云博客

本文介绍了一种使用深度优先搜索(DFS)算法来解决迷宫中最远可达距离问题的方法。通过递归地探索迷宫中所有可能路径，并标记已访问位置的距离，最终找出迷宫中可达的最远距离。

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/*
ID: niepeng1
LANG: C++
TASK: maze1
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <iostream>
using namespace std;

struct Node{
int i;
int j;
int num;
};
char map[202][80];
int map2[201][80];
int w,h,w2,h2,max2=0;
int point[2][2];

void Dfs(int i,int j,int count)
{
if( map[i][j-1] != '|' && (map[i][j-2] == ' ' || map2[i][j-2] > count+1) ){
  map[i][j-2] = 'a';
  map2[i][j-2] = count+1;
  Dfs(i,j-2,count+1);
}
if(map[i][j+1] != '|' && (map[i][j+2] == ' ' || map2[i][j+2] > count+1) ){
   map[i][j+2] = 'a';
   map2[i][j+2] = count+1;
   Dfs(i,j+2,count+1);
}
if(map[i-1][j] != '-' && (map[i-2][j] == ' ' || map2[i-2][j] > count+1) ){
   map[i-2][j] = 'a';
   map2[i-2][j] = count+1;
   Dfs(i-2,j,count+1);
}
if(map[i+1][j] != '-' && (map[i+2][j] == ' ' || map2[i+2][j] > count+1) ){
   map[i+2][j] = 'a';
   map2[i+2][j] = count+1;
   Dfs(i+2,j,count+1);
}
}
int main(void)
{
int i,j;
char c;
FILE *in, *out;
in = fopen("maze1.in", "r");
out = fopen("maze1.out", "w+");

fscanf(in,"%d",&w); w2 = 2*w;
fscanf(in,"%d",&h); h2 = 2*h;
fscanf(in,"%c",&c);
for(i=0; i<=h2; i++)
{
  for(j=0; j<=w2; j++)
  {
   map2[i][j]=0;
   fscanf(in,"%c",&map[i][j]);
  }
  fscanf(in,"%c",&c);
}
for(j=0,i=0;i<=h2;i++){
  if(map[i][0] == ' '){
   point[j][0] = i;
   point[j][1] = 1;
   map[i][0] = '|';
   j++;
  }
  if(map[i][w2] == ' '){
   point[j][0] = i;
   point[j][1] = w2-1;
   map[i][w2] = '|';
   j++;
  }
}
for(i=0;i<=w2;i++){
  if( map[0][i] == ' '){
   point[j][0] = 1;
   point[j][1] = i;
   map[0][i] = '-';
   j++;
  }
  if(map[h2][i] == ' '){
   point[j][0] = h2-1;
   point[j][1] = i;
   map[h2][i] = '-';
   j++;
  }
}
map2[ point[0][0] ][ point[0][1] ] =1;
map2[ point[1][0] ][ point[1][1] ] =1;
Dfs(point[0][0],point[0][1],1);
Dfs(point[1][0],point[1][1],1);

for(i=1;i<=h2-1;i+=2)
  for(j=1;j<=w2-1;j+=2)
  {
   if( max2 < map2[i][j])
    max2=map2[i][j];
  }
fprintf(out,"%d/n",max2);//out,
//printf("%d/n",max2);
fclose(in);
fclose(out);

return 0;
}